A hole of area 2.0 cm² at the bottom of a tank 2.0 m deep is closed with a cork. Determine the force on the cork when the tank is filled with water. [ Density of water is 1000kg/m³ and acceleration due to gravity is 10m/s²] [4mks]

In the set up shown in the fig below, water near the top of the boiling tube boils while at the bottom remains cold.

Give a reason for the observation. [2mks]

…………………………………………………………………………………………

State three ways of increasing the sensitivity of a liquid – in – glass thermometer. [3mks]

…………………………………………………………………………………………

.i) Define friction. [1mks]

…………………………………………………………………………………………

ii) A wooden box of mass 30kg rests on a rough floor. The co –efficient of friction between the floor and the box is 0.6.

Calculate the force just required to just move the box. [3mks]

iii) State three methods of minimising friction. [3mks]

…………………………………………………………………………………………

.i) State two factor that affect conduction of heat in solids. [2mks]

…………………………………………………………………………………………

ii) Two identical aluminium rods are shown in the figure below. One rests on metal block and the other on a wooden block. The protruding ends are heated on a Bunsen burner as shown.

State with reason on which bar the wax is likely to melt sooner. [2mks]

…………………………………………………………………………………………

SECTION B. (55MKS)

A man uses an inclined plane to lift a 50kg load through a vertical height of 4m. The inclined plane makes an angle of 30˚ with the horizontal. If the efficiency of the inclined plane is 72%. Calculate
The effort needed to move the load up the inclined plane at a constant velocity. [3mks]

The work done against friction in raising the load through a height of 4.0 m.

[Take g = 10N /kg] [3mks]

A certain machine raises 2.0 tonnes of water through 22m. If the efficiency of the machine is 80%, how much work is done by the machine? [3mks]

(a) i) The fig below shows air flowing through a pipe of different cross-section areas. Two pipes A &B. are dipped into water.

Explain the cause of difference in the levels of water in the pipes A and B. [2mks]

…………………………………………………………………………………………

ii)The fig below shows Bunsen burner.

Explain how air is drawn into the burner when the gas tap is opened. [3mks]

…………………………………………………………………………………………

i) What is meant by the centre of gravity of a body? [1mk]

…………………………………………………………………………………………

ii) The figure below is at equilibrium.

Given that the meter rule is uniform, determine its weight. [5mks]

(a)Explain why a drop of methylated spirit on the back of the hand feels colder than a drop of water at the same temperature. [2mks]

…………………………………………………………………………………………

b) State the meaning of the term ‘specific latent heat of fusion’ [1mks]

………………………………………………………………………………………

The fig below shows a setup of apparatus set in an experiment to determine the specific latent heat of fusion.

The following reading was noted after the heater was switched on for 5 minutes.

Mass of beaker = 130g
Mass of beaker + melted ice = 190g

Determine the,
Energy supplied by the 60W heat in 5 minutes. [3mks]

Specific latent heat of fusion of ice. [4mks]

It was observed that some of the crushed ice melted even before the heater was switched on. State the reason for this observation. [1mk]

…………………………………………………………………………………………

Brownian motion of smoke particles can be studied by using the apparatus shown in the figure below. To observe the motion some smoke is enclosed in the smoke cell and then observed through the microscope..

Explain the role of the
Smoke particles [2mks]

…………………………………………………………………………………………

Lens [2mks]

…………………………………………………………………………………………

Microscope [2mks]

…………………………………………………………………………………………

State and explain the nature of the observed motion of the smoke particles. [3mks]

…………………………………………………………………………………………

State what will be observed about the motion of the smoke particles if the temperature surrounding the smoke cell is raised slightly. [1mks]

…………………………………………………………………………………………

(a) The fig below shows a velocity – time graph for the motion of a certain body.

Describe the motion of the body in the region. [3mks]

…………………………………………………………………………………………

b) A car initially at 10m/s decelerates at2.5m/s².

Determine,

Its velocity after 1.5 s [2mks]

The distance travelled in 1.5 s. [2mks]

III. The time taken for the car to stop. [2mks]

(c)

State Newton’s first law of motion. [1mks]

…………………………………………………………………………………………

A stone of mass 0.01kg is released from a catapult of mass 0.25kg. If the release velocity of the stone is 175m/s, determine the recoil velocity of the catapult. [4mks]

___________________________________________________________________

PHYSICS PAPER 1 FORM 3 MARKING SCHEME

Volume = 7.4cm³ – 4.6cm³ = 2.8 cm³

Density = mass = 11g = 3.9/cm³

Volume 2.8cm³

In (a) the adhesive force between water and capillary tube is stronger than the cohesive force between water molecules while in (b) the cohesive force between mercury molecules is stronger than adhesive force between mercury and the tube.
Force = Pressure x Area

P=hpg

=2x1000x10= 20 000

A = 2

10000

=0.0002m²

P = FxA

= 20000x 0.0002 = 4N

Water and glass are poor conductors of heat.
(i)Making the bore very fine

ii)Reducing the size of the bulb

iii)Making the wall of the bulb very thin

(i)Friction is a force that resists one surface from sliding over the other.
ii) F = R

= mg

=0.6 x 30×10

= 180N

iii)- Use of rollers

Use of ball bearings
Lubrication

(i)- Length of a conductor

Cross-section area of a conductor
Temperature difference between the ends of a conductor
The nature of the material. [any two]

ii)On the wooden block. The wooden block is a poor conductor of heat and so all the heat goes in melting the wax.

SECTION B

V.R = 1

SIN 30

= 2

M.A= efficiency x V.R

= 72 X 2

100

= 1.44

Effort = load

M.A

= 50×10

1.44

= 347.2N

b)Work done against friction = work input – work output

work out put = mgh

= 50x10x4 = 2000J

Work input = effort x distance moved by effort

= 347.2 x AC

= 347.2 X 4

SIN 30

= 2777.65

Therefore work done against friction

= 2777.6 – 2000

= 777.6J

c) Efficiency = work output x 100

Work input

Work input = 2000×10 x 22 = 550 000J

0.8

(a)i)The pressure above tube B is low than that above tube A. this is caused by high velocity of air above B compared to velocity above tube A.
ii) When the gas tap is opened, gas flows at high speed creating a low pressure region above the nozzle. The higher atmospheric pressure on the outside pushes air in and the gas burns
b) i) Centre of gravity is the point of application of resultant force due to earth’s attraction on the body.
ii) Sum of clockwise moment = sum of anticlockwise moments

0.1xw + (1×0.60) = 2×0.4

0.1w + 0.6 = 0.8

0.1w = 0.8 -0.6

0.1w = 0.2

W = 2N

(a)Methylated spirit evaporates faster [is highly volatile] than water taking away the latant heat faster from the hand.

b)Specific latent heat of fusion of a substance is the quantity of heat required to melt completely one kilogram of a substance at constant temperature.

c)i)I. E = pt

= 60x5x60 = 18000J

E = L_f

18000 = 60 x L_f

1000

L_f= 300000J/kg

ii)The temperature of the room melted some ice since it was higher than the melting point of ice.

(a)i)Smoke particles are used to show the behavior of air molecules since they are more visible than air molecules and light enough to move when bombardment by air molecules.
ii) The lens focuses the light from the lamp on the smoke particles causing them to be observable.

iii)The microscope enlarges [magnifies] the smoke particles so that they are visible.

b)The smoke particles are seen to be in constant random motion. The motion is caused by uneven bombarded by invisible particles of air.

c) The smoke particles would move faster.

(a)i) OA : Body moves with constant acceleration or the body moves with uniformly increasing velocity.

ii)AB: Body moves with non-uniform decceleration.

iii)BC : Body moves with constant velocity or moves with zero acceleration.

b)i)I. V = U + at

= 10-(2.5×1.5)

= 6.25m/s

S= Ut + ½ at²

=(10×1.5)- ( ½ x2.5 x 1.5²)

= 12.1875m

III. V= u +at

O = 10 – 2.5t

T = 4 seconds

c)i) A body remains at rest or in a uniform motion in a straight line, unless acted upon by an external force.

ii)Total momentum before releasing the catapult.

= m₁u₁ + m₂u₂

=(0.25 x 0 )+ (0.01 x 0)

[Both catapult and stone at rest]

= 0kgm/s

Momentum of the catapult after release

=0.25 x v

=0.25vkgm/s

Momentum of the stone after release

= 0.01 x 175

=1.75kgm/s

Total final momentum is given by

= 1.75 + 0.25v

Initial momentum = final momentum

O = 1.75 + 0.25v

-1.75=0.25v

-1.75 = V

0.25

– 7 = V

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